Question: Let $a(x)=-5x^3-x^2+3$, and $b(x)=x^2+4$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{-5x^3-x^2+3}{x^2+4}$ : First, we divide ${x^2}$ into ${-5x^3}$ and get ${-5x}$ : $ \hphantom{1567|14} {-5x}\\ {{{x^2}+4}}|\overline{{-5x^3}-x^2\ +\ 0x+\ 3}\\ \hphantom{37....|}\llap{-}\underline{(-5x^3+0x^2-20x)}\\ \hphantom{37|3.............}-x^2+\ 20x\\ $ [What did we do here?] Next, we divide ${x^2}$ into ${-x^2}$ to get ${-1}$ : $ \hphantom{1567|14} {-5x \ {- \ 1}}\\ {{{x^2}+4}}|\overline{-5x^3-x^2\ +\ 0x\ +3}\\ \hphantom{37....|}\llap{-}\underline{(-5x^3+0x^2-20x)}\\ \hphantom{37|3.............}{-x^2}+\ 20x+3\\ \hphantom{37...............|}\llap{-}\underline{(-x^2\ +\ 0x\ - 4)}\\ \hphantom{37|3.....................}{+20x\ +\ 7}\\ $ [What did we do here?] The process stops here because $x^2+4$ is a polynomial of the second degree and $20x+7$ is a polynomial of the first degree. So it follows that ${r(x)}={20x+7}$, ${q(x)}={-5x-1}$, and $ \dfrac{-5x^3-x^2+3}{x^2+4}={-5x-1}+\dfrac{{20x+7}}{x^2+4}$ To conclude, $q(x)=-5x-1$ $r(x)=20x+7$